從PHP文件接收HTTP POST回顯響應（發送POSTS效果很好，這是我不知道的接收信息）

Question

因此，正如標題所示，我的問題是得到我正在發出的HTTP POST的響應。 應該發生的是，我發送了一堆變量，PHP檢查數據庫中的變量並將結果發送回給我（作為對頁面的回顯）。

這是android代碼：

 public class CheckChallenge extends AsyncTask<String, Void, String> 
  {
    @Override
    protected String doInBackground(String... urls) 
    {
      String response = "";
      try
      {
        URL = urls[0];
   ArrayList<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>(4);
   nameValuePairs.add(new BasicNameValuePair("victim",NetPlay.myId)); 
   // need to return these to an array
   nameValuePairs.add(new BasicNameValuePair("rival",rivalid));


        nameValuePairs.add(new BasicNameValuePair("word","null"));
        nameValuePairs.add(new BasicNameValuePair("won","0"));

          HttpClient httpclient = new DefaultHttpClient();
          HttpPost httppost = new      
          HttpPost("http://www.hanged.comli.com/check-rival.php");
          httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));

          HttpResponse execute = httpclient.execute(httppost);
          HttpEntity entity = execute.getEntity();

          //InputStream is = entity.getContent();

          //mText.setText(is.toString());

         Log.i("postData", execute.getStatusLine().toString());
         //HttpEntity entity = response.getEntity();

      }
      catch(Exception e)
      {
              Log.e("log_tag", "Error in http connection"+e.toString());
      }
            return response;
        } 

    @Override
    protected void onPostExecute(String result) 
    {
        // CHECK ENTIRE DATABASE FOR MY ID //
        // IF MY ID IS THERE THEN THAT MEANS IVE CHALLENGED SOMEONE //


    }

  }

這是我認為可以的PHP，僅出於完整性考慮就可以了：$ connect = mysql_connect（“ $ mysql_host”，“ $ mysql_user”，“ $ mysql_password”）或die（“ cannot connect”）; mysql_select_db（“ $ mysql_database”，$ connect）或die（“無法選擇數據庫”）; session_start（）;

$victim = $_POST['victim'];
$rival = $_POST['rival'];
$word = $_POST['word'];
$won = $_POST['won'];

$query = "SELECT rival FROM currentgames";
$result = mysql_query($query);

 if (!$result) 
 {
echo "Could not successfully run query ($sql) from DB: " . mysql_error();
exit;
}

if (mysql_num_rows($result) == 0) 
{
echo "No rows found, nothing to print so am exiting";
exit;
}

 while ($row = mysql_fetch_assoc($result)) 
{
 echo $row["rival"];
}

對此的任何幫助將不勝感激，嘗試使我了解所有這些HTTP POSTing內容。

Answer 1

發送HTTP請求並回讀HTTP響應的示例：

String res = "";
String url = "http://www.domain.com/youscript.php";
URL urlObj = new URL(url);
URLConnection lu = urlObj.openConnection();


// Send data - if you don't need to send data 
// ignore this section and just move on to the next one
String data = URLEncoder.encode("yourdata", "UTF-8");
lu.setDoOutput(true);
OutputStreamWriter wr = new OutputStreamWriter(lu.getOutputStream());
wr.write(data);
wr.flush();

// Get the response
BufferedReader rd = new BufferedReader(new InputStreamReader(lu.getInputStream()));
String line = "", res = "";
while ((line = rd.readLine()) != null) {
  res += line;
}

wr.flush();
wr.close();
System.out.println(res);